Automata and nested recurrences Jeffrey Shallit School of Computer Science University of Waterloo Waterloo, Ontario N2L 3G1 Canada shallit@cs.uwaterloo.ca http://www.cs.uwaterloo.ca/~shallit Joint work with Jean-Paul Allouche. 1 / 22
Hofstadter’s recurrence Douglas Hofstadter, G¨ odel, Escher, Bach , 1979: Q ( n ) = Q ( n − Q ( n − 1)) + Q ( n − Q ( n − 2)) for n ≥ 2 and Q (1) = Q (2) = 1. 2 / 22
1st differences of Hofstadter’s Q-sequence Larger and larger hedgehogs 3 / 22
Hofstadter and Huber Hofstadter and Huber (1999): Q r , s ( n ) = Q r , s ( n − Q r , s ( n − r )) + Q r , s ( n − Q r , s ( n − s )) for n > s > r . Balamohan, Kuznetsov and Tanny (2007): a nearly complete analysis of the sequence Q 1 , 4 (called V in their paper). It is defined by V (1) = V (2) = V (3) = V (4) = 1 , and ∀ n > 4 , V ( n ) := V ( n − V ( n − 1)) + V ( n − V ( n − 4)) . 4 / 22
The V-sequence n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 V ( n ) 1 1 1 1 2 3 4 5 5 6 6 7 8 8 5 / 22
The V -sequence ◮ is monotone increasing ◮ successive terms differ by 0 or 1 ◮ no number appears more than four times (and only 1 appears that many) ◮ grows approximately like n / 2 V ( n ) − n / 2 6 / 22
Frequency sequence n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 V ( n ) 1 1 1 1 2 3 4 5 5 6 6 7 8 8 Balamohan, Kuznetsov, and Tanny (2007): a precise description of the “frequency” sequence F ( n ) defined by F ( a ) := # { n : V ( n ) = a } . n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 F ( n ) 4 1 1 1 2 2 1 2 2 1 3 2 1 2 7 / 22
Explicit description Theorem (Balamohan, Kuznetsov, Tanny) There exist two (explicit) maps g , h, with g , h : { 1 , 2 , 3 } 4 → { 1 , 2 , 3 } , such that, for all a > 3 F (2 a ) = g ( F ( a − 2) , F ( a − 1) , F ( a ) , F ( a + 1)) F (2 a + 1) = h ( F ( a − 2) , F ( a − 1) , F ( a ) , F ( a + 1)) . 8 / 22
Corollary of our main result Theorem. (Allouche & JOS, 2012) The sequence ( F ( n )) n ≥ 1 is 2-automatic. ◮ This means that F ( n ) can be computed “in a simple way” from the base-2 representation of n . ◮ In particular, it can be computed in O (log n ) time. ◮ Furthermore, we can compute the automaton explicitly. 9 / 22
What is an automatic sequence? ◮ An infinite sequence a = a 0 a 1 a 2 · · · over a finite alphabet of letters, generated by a finite-state machine (automaton) ◮ The automaton, given n as input, computes a n as follows: ◮ n is represented in some fixed integer base k ≥ 2 ◮ The automaton moves from state to state according to this input ◮ Each state has an output letter associated with it ◮ The output on input n is the output associated with the last state reached 10 / 22
The canonical example: the Thue-Morse automaton 0 0 1 0 1 1 This automaton generates the Thue-Morse sequence t = ( t n ) n ≥ 0 = 0110100110010110 · · · . 11 / 22
Why automatic sequences? ◮ A nontrivial class of self-similar sequences ◮ Many “naturally-occurring” sequences are automatic ◮ Halfway between periodic and chaotic ◮ Provide canonical examples for various kinds of avoidance problems ◮ Statistics like the limiting frequency of occurrence of any letter are easy to compute 12 / 22
The main result Theorem. If a sequence F satisfies a recurrence “like” F (2 a ) = g ( F ( a − 2) , F ( a − 1) , F ( a ) , F ( a + 1)) F (2 a + 1) = h ( F ( a − 2) , F ( a − 1) , F ( a ) , F ( a + 1)) . then F is 2-automatic. Proof idea. Find a finite class of sequences of which F is a member, and which is closed under the transformations n → 2 n ; n → 2 n + 1 . The proof is explicit, but the automaton it constructs in our case is too large to deal with. 13 / 22
Constructing the automaton We constructed the automaton by “guessing” it and checking it. The states are named by binary strings. We guess that x and y correspond to the same state if F ([ xz ]) and F ([ yz ]) are the same for all the z we can reasonably test (say | z | ≤ 20). Actually, this is not quite good enough; we actually deal with the 4-tuple ( F ([ x ] − 2) , F ([ x ] − 1) , F ([ x ]) , F ([ x ] + 1)) . This gives us a 33-state automaton that we can now verify by a very tedious induction on | x | , using the recurrence we know. Now, throwing away all but the 3rd of 4 outputs for each state, we get a smaller automaton by minimization. 14 / 22
A conjecture Conjecture. V ′ , the first difference sequence of V ( n ), is also 2-automatic. For example, up to the limit we have checked (tens of thousands of terms) V ′ ([10110 x ]) V ′ ([10001 x ]) = V ′ ([11000 x ]) V ′ ([10011 x ]) = V ′ ([100011 x ]) V ′ ([11110 x ]) = V ′ ([101010 x ]) V ′ ([100000 x ]) = V ′ ([101011 x ]) V ′ ([100001 x ]) = V ′ ([101110 x ]) V ′ ([100100 x ]) = V ′ ([101111 x ]) V ′ ([100101 x ]) = However, if it is 2-automatic, then the smallest automaton (msd first) has at least 900 states. 15 / 22
Morphic sequences: a generalization of automatic sequences A morphism is a map h from a finite alphabet Σ ∗ → ∆ ∗ satisfying h ( xy ) = h ( x ) h ( y ) for all strings x , y . If Σ = ∆ we can iterate h . If h ( a ) = ax for a letter a then h ω ( a ) = a x h ( x ) h 2 ( x ) . . . . Note that h ω ( a ) is a fixed point of h . If a sequence arises by applying a coding (a renaming of the letters) to h ω ( a ) then it is called morphic . 16 / 22
Slow sequences Recently Allouche has proposed studying “slow” sequences, that is, sequences of natural numbers where the first differences are either 0 or 1. Theorem. Let x = ( x n ) n ≥ 1 be a slow sequence. Then the corresponding frequency sequence F ( x ) is bounded and morphic iff the the first difference sequence ∆ x is morphic and 1’s appear there with bounded gaps. Corollary The first difference sequence of V is morphic. 17 / 22
Proof of the theorem i − 1 � �� � Proof. Apply the morphism that sends i to 1 0 0 · · · 0 to the frequency sequence. Example: original sequence x is 1 1 1 1 2 3 4 5 5 6 6 78 8 9 9 10 11 11 11 · · · Frequency sequence is 41112212213 · · · Apply morphism to that to get 10001111010110101100 · · · , which, except for the first term, is ∆ x . 18 / 22
Proof of the other direction For the other direction, apply a finite-state transducer that measures the distance between consecutive 1’s: Example: if the first difference sequence is (1)00011110101101011001 · · · , then the transducer outputs 41112212213 · · · . However, these implications need not hold for k -automatic sequences... 19 / 22
Example 1 Suppose the frequency sequence F ( x ) = ( f ( n )) n ≥ 1 is given by � 2 , if n is one less than a power of 2; f ( n ) = 1 , otherwise. So ( f ( n )) n ≥ 1 is 21211121111111211111111111111121 · · · Then x is 1 1 2 3 3 4 5 6 7 7 8 9 10 11 12 13 14 15 15 16 17 · · · and the first difference of this is 01101111011111111011111111111111 · · · with 0’s in positions 1 4 9 18 35 68 · · · and generally in positions 2 n + n − 2. This is not 2-automatic. 20 / 22
Example 2 On the other hand if we define the first difference sequence by � 0 , if n is a power of 2; g ( n ) = 1 , otherwise. Then ( g ( n )) n ≥ 1 is automatic and equals 00101110111111101111 · · · so the original sequence is 1 1 1 2 2 3 4 5 5 6 7 8 9 10 11 12 12 13 14 15 16 17 18 19 · · · and the corresponding frequency sequence is 321121111112111111111111112 · · · with 2’s in positions 2 5 12 27 58 · · · and in general 2 n − n , so this is not automatic either. 21 / 22
For further reading 1. J.-P. Allouche and J. Shallit, A variant of Hofstadter’s sequence and finite automata, J. Aust. Math. Soc. 93 (2012), 1–8. 2. B. Balamohan, A. Kuznetsov, and S. Tanny, On the behavior of a variant of Hofstadter’s Q -sequence, J. Integer Seq. 10 (2007) Article 07.7.1. 3. K. Pinn, Order and chaos in Hofstadter’s Q ( n ) sequence, Complexity 4 (1999), 41–46. 22 / 22
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